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Let $(M,g)$ be a Riemannian manifold and $V$ a unit Killing vector field on it. Under what condition on curvature tensor the following equation hold: $$\nabla_VQ=0,$$ where $Q$ is the Ricci operator defined as $g(QX,Y)=\rho(X,Y)$.

Update1: Einstein metrics that admit a unit Killing vector field satisfies the above equation.

Update2: Can anybody give an example other than Einstein manifolds that satisfy the above equation?

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  • $\begingroup$ Cartesian product of a sphere with a hyperbolic manifold. $\endgroup$ – Deane Yang Aug 3 '17 at 13:35
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If $V$ is a Killing vector, then necessarily its Lie derivative annihilates all curvature tensors, including the Ricci operator, namely $$ \mathcal{L}_V Q = \nabla_V Q + Q J_V - J_V Q = 0 , $$ where $J_V$ is the ("Jacobian") endomorphism on tangent vectors defined by $\nabla_U V = J_V U$. Thus, the condition that you want seems to be just $Q J_V - J_V Q = 0$. I don't see how the unit condition on $V$ enters.

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  • $\begingroup$ Can be obtained this condition on Riemann curvature tensor? $\endgroup$ – C.F.G Aug 2 '17 at 17:22
  • $\begingroup$ Yes. Another way to think about this is that a Killing vector field is just an infinitesimal variation of a family of isometries. But the pull back of the metric under the isometry is just the metric itself, so the pullback of any local invariant, such as the curvature tensor and its covariant derivatives will be equal to the original value. Therefore, the infinitesimal variation has to be zero. $\endgroup$ – Deane Yang Aug 2 '17 at 20:15
  • $\begingroup$ @C.F.G I'm not sure I understand your question. A condition on the Ricci operator is by definition also a condition on the Riemann tensor. Maybe you are looking for a condition independent of $V$? $\endgroup$ – Igor Khavkine Aug 3 '17 at 4:12
  • $\begingroup$ You are right. I want to know if we assume that $(M,g)$ is Einstein manifold then can be determine curvature tensor completely? Answer of my second update is positive or not? $\endgroup$ – C.F.G Aug 3 '17 at 6:05

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